Question: Divide the following rational expressions and simplify the result. $\dfrac{2xz^2+7x^2z}{12xz^2+36x^2z-120x^3}\div\dfrac{12z^2+42xz}{x^3z+5x^4}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $2xz^2+7x^2z$, of the dividend can be factored as $xz(2z+7x)$ by factoring out $xz$. The denominator, $12xz^2+36x^2z-120x^3$, of the dividend can be factored as $12x(z+5x)(z-2x)$ by factoring out $12x$ and using the sum-product pattern. The numerator, $12z^2+42xz$, of the divisor can be factored as $6z(2z+7x)$ by factoring out $6z$. The denominator, $x^3z+5x^4$, of the divisor can be factored as $x^3(z+5x)$ by factoring out $x^3$. Now the division looks as follows: $\dfrac{xz(2z+7x)}{12x(z+5x)(z-2x)}\div\dfrac{6z(2z+7x)}{x^3(z+5x)}$ To divide two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{xz(2z+7x)}{12x(z+5x)(z-2x)}\div\dfrac{6z(2z+7x)}{x^3(z+5x)} \\\\\\ &= \dfrac{xz(2z+7x)}{12x(z+5x)(z-2x)}\cdot \dfrac{x^3(z+5x)}{6z(2z+7x)} &\text{Flip the divisor.}\\\\\\\\ &= \dfrac{xz(2z+7x) \cdot x^3(z+5x)}{12x(z+5x)(z-2x) \cdot (6z(2z+7x))} &\text{Multiply across.}\\\\\\\\ &= \dfrac{{\cancel{x}}{\cancel{z}}{\cancel{(2z\!+\!7x)}} x^3{\cancel{(z\!+\!5x)}}}{12{\cancel{x}}{\cancel{(z\!+\!5x)}}\!(z\!-\!2x)6{\cancel{z}}\!{\cancel{(2z\!+\!7x)}}} &\text{Cancel out common factors.}\\\\\\\\ &= \dfrac{x^3}{72(z-2x)} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{x^3}{72(z-2x)}$, which is equivalent to $\dfrac{x^3}{72z-144x}$.